Problem 74 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{x \rightarrow 0} \frac{\left(2^{m}+x\right)^{1 / m}-\left(2^{n}+x\right)^{1 / n}}{x}\) is equal to (A) \(\frac{1}{m 2^{m}}-\frac{1}{n 2^{n}}\) (B) \(\frac{1}{m 2^{m}}+\frac{1}{n 2^{n}}\) (C) \(\frac{1}{m 2^{m-1}}-\frac{1}{n 2^{n-1}}\) (D) None of these

Step-by-Step Solution

Answer

Option (C): \( \frac{1}{m 2^{m-1}} - \frac{1}{n 2^{n-1}} \).

Step 1: Simplify the Expression

We begin by rewriting the expression in the limit: \( \lim _{x \rightarrow 0} \frac{(2^{m}+x)^{1/m}-(2^{n}+x)^{1/n}}{x} \). Notice that this expression resembles a derivative form, \( \frac{f(x) - f(a)}{x-a} \), where \( f(x) = (a + x)^{1/p} \). Simplify it considering \( x \to 0 \).

Step 2: Apply L'Hopital's Rule

The limit \( \lim _{x \to 0} \frac{(2^{m}+x)^{1/m}-(2^{n}+x)^{1/n}}{x} \) is an indeterminate form \( \frac{0}{0} \). Apply L'Hopital's Rule, which allows differentiation of the numerator and the denominator. Differentiate \( (2^{m}+x)^{1/m} \) as \( \frac{d}{dx}(2^{m}+x)^{1/m} = \frac{1}{m}(2^{m}+x)^{1/m - 1} \) and similarly for \( (2^{n}+x)^{1/n} \).

Step 3: Differentiate the Numerators

Differentiate \( (2^{m}+x)^{1/m} \) to get \( \frac{1}{m}(2^{m}+x)^{1/m -1} \) and differentiate \( (2^{n}+x)^{1/n} \) to get \( \frac{1}{n}(2^{n}+x)^{1/n -1} \).

Step 4: Calculate the Limit

After differentiation, calculate the limit by substituting \( x=0 \) into the new expression: \( \lim_{x \to 0} \left( \frac{1}{m}(2^{m}+x)^{1/m -1} - \frac{1}{n}(2^{n}+x)^{1/n -1} \right) \). This simplifies to \( \frac{1}{m 2^{m-1}} - \frac{1}{n 2^{n-1}} \).

Step 5: Determine the Correct Option

Evaluate the expression \( \frac{1}{m 2^{m-1}} - \frac{1}{n 2^{n-1}} \) and compare with the given options. The correct answer is option (C): \( \frac{1}{m 2^{m-1}} - \frac{1}{n 2^{n-1}} \).

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a useful tool in calculus for solving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you find a limit that gives you these forms, you'll want to use L'Hopital's Rule. Here's how it works:

• Differentiation: You differentiate the numerator and the denominator separately.
• Re-evaluate: Substitute back the limit into this new expression to see if you can find a value.
• Repeat: If it still results in an indeterminate form, apply L'Hopital's Rule again.

In our example, we had the expression \( \lim_{x \to 0} \frac{(2^{m}+x)^{1/m}-(2^{n}+x)^{1/n}}{x} \), which was \( \frac{0}{0} \) at \( x=0 \). By differentiating both the numerator and the denominator, L'Hopital's Rule helps us find the limit step-by-step.To apply L'Hopital's Rule effectively, it is important to first confirm that the function is indeed indeterminate and to do this calculation accurately, or else the result might be incorrect.

Indeterminate Forms

Indeterminate forms are expressions in calculus that do not have a well-defined limit. This usually includes forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others such as \( \infty - \infty, 0 \times \infty \), etc. The \( \frac{0}{0} \) form occurs in our exercise and is crucial for applying L'Hopital's Rule.For our problem:

• The original function \( \frac{(2^{m}+x)^{1/m}-(2^{n}+x)^{1/n}}{x} \) at \( x=0 \) gives \( \frac{0}{0} \), indicating that it's an indeterminate form.
• Recognizing this form is the first step to deciding to use L'Hopital's Rule.

By transforming this indeterminate form using derivatives, we can find a limit that avoids the ambiguity of forms like \( \frac{0}{0} \).Understanding indeterminate forms help you know exactly when and how to apply strategies, like L'Hopital's Rule, to find true limit values.

Derivative Calculation

Calculating derivatives is central to applying L'Hopital's Rule. It's all about finding rates of change. Here's a breakdown on how we use it in this exercise:

• The derivative of \( (2^{m}+x)^{1/m} \) is found using the power rule and chain rule, giving \( \frac{1}{m}(2^{m}+x)^{1/m -1} \).
• Similarly, for \( (2^{n}+x)^{1/n} \), the derivative is \( \frac{1}{n}(2^{n}+x)^{1/n -1} \).
• You apply these derivatives to the original indeterminate form to simplify the expression.

Once you have these derivatives, you can substitute back \( x=0 \) to compute the limit. The derivative calculation is key because it turns the complex problem of limits into a straightforward arithmetic problem.This process not only helps in this specific exercise but also builds a foundation for tackling more complex calculus problems."}]}]}​​